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Awesome! Thank you

Thanks guys that really helped!

i can use websites if i have the url only 

-4√(x - 3)   < - 8      divide both sides by  - 4   and reverse the inequality sign

√[x - 3]  >  2         square both sides

x - 3  >  4            add 3 to both sides

x >  7

And you have it graphed correctly  !!!!!

We want to find  a  such that the numerator is also zero when  x = 3

That is, we want to find  a  such that...

2(3)² - 3a(3) + 3 - a - 1  =  0

18 - 9a + 3 - a - 1  =  0

20 - 10a  =  0

20  =  10a

2  =  a

So yes, when  a = 2,  the limit exists.

Check:

\(\lim\limits_{x\to3}\frac{2x^2-6x+x-3}{x^2-2x-3}\ =\ \lim\limits_{x\to3}\frac{2x(x-3)+1(x-3)}{(x-3)(x+1)}\ =\ \lim\limits_{x\to3}\frac{2x+1}{x+1}\ =\frac74\)

what school do you go to bc i know the websites that arent blocked for other schools.

Exactly   ....!!!

If   f(a)   produces anything but a "0"....then "a"  is not a root  and  (x - a)  is not a factor....

thank you!

Awesome, so I inputted it correct?

Manu evaluates  f(-4)  and determines its value to be  -4 . Manu concludes that x + 4 is not a factor of f(x)=−x^4−6x^3+15x−72 .

Like this ?

Positive from  (-infinity, -2) and  (8, infinity)

Negative from  (-2, 8)

Manu needs to determine whether x + 4 is a factor of f(x)=−x^4−6x^3+15x−72 .

How can Manu use the factor theorem to determine whether x + 4 is a factor of f(x) ?

If x + 4 is a factor, then x = -4  is a root.....use synthetic division....if we get a 0, then x + 4  is a factor....so

-4  [ -1       -6         0         15       -72  ]

                  4          8        -32        68

     ____________________________

       -1      -2          8        -17        -4

We have a remainder of -4, so..  x = -4  is not a root and ..x + 4  is NOT a factor

(x+4) is not.

One  way is to  convert  (2, pi/6)   into  Cartesian Coordinates.....so we have

To

(2cos (pi/6) , 2sin (pi/6) )   =   (2 * √3/2, 2 * 1/2)   = (√3/2 , 1 )

One possible point for the remaining vertex we have

(√3/2  + 0, 1 + 0)  -  (4,0)   =  (√3/2 - 4, 1)

The area of this parallelogram  (ABCD)  =   base * height  =  4 * 1  =  4 units^2

Another possible vertex  is given by

(√3/2 + 4, 1 + 0)  - (0, 0)   =  (√3/2 + 4, 1)

The area of this parallelogram (ABCE) is the same as the first one

The last possible vertex is

(4 + 0, 0 + 0)  - (√3/2 , 1)   =   ( 4 - √3/2, -1)

And the area of this paralleogram ( AFBC) = area of triangle ABC + area of triangle ABF  =   4 units^2

So all the possible parallelograms have the same areas  !!!

See the following image  :

https://www.chegg.com/homework-help/explore-cases-0-upper-bound-lower-bound-real-zeros-polynomia-chapter-4.2-problem-69e-solution-9780073519517-exc

If you click on this link the upper boun therm states "by using synthetic divsion if alll numbers are postitive and then thats the upper bound "

When i solve 7 into the equation so 7/4 -25 -5 -13 , i got all postitive values and then once i do 6 it is a mis of pos and neg values

I don't quite understand what you mean.......we are looking for the possible upper bound on zeroes.....in other words.....we are trying to find the largest x value such that no positive root can exceed this upper bound 

Look at the graph here  :  https://www.desmos.com/calculator/omunuy8c4t

Note that the largest real root occurs at about (2.68, 0).....so   this confirms the fact that we do not need to search for any real roots  > 7.25 because they do not exist

I think that the  inequality  is incorrect

See the graph here :  https://www.desmos.com/calculator/alijngoclf

Notice that x^2   is  ≥ sin^2x    everywhere on   (-pi/2, pi/2)

But  if     x^2  ≤ f(x)   on this interval...then f(x) is ≥ x^2  everywhere on this interval...then f(x)  cannot also be  ≤ sin^2x  everywhere on the interval

I believe that the inequality we need is  this

sin^2 x  ≤ f(x)   ≤ x^2

Note that if this true......then

f(x)                 f(x)      =    0          by the Squeeze Theorem

lim x → 0   

Is it possible for it to be 7 ? Because when I didive 7 by the polyonmial I get all postitive values and that is lower than 7.25?

Thanks so much!! 

   x                 20                             2

___  =       ___________   +    _______

x - 4          (x + 4) ( x - 1)             x  - 1

Multiply through by the common denominator   ( x - 4) ( x + 4) ( x - 1)    and we have

x (x + 4) ( x - 1)  =  20 (x - 4)  +  2 (x - 4) (x + 4)      simplify

x ( x^2 + 3x - 4)  =  20x - 80 + 2( x^2 - 16)

x^3 + 3x^2 - 4x   =  20x - 80  + 2x^2 - 32

x^3 + x^2 - 24x + 112 =  0    

WolframAlpha shows that the only real solution  is   x  ≈ -6.8684

https://www.wolframalpha.com/input/?i=x%5E3+%2B+x%5E2+-+24x+%2B+112++%3D++0

Glad to see you here Rosala! Hows the computer now?

im doing good now that my computer is back aliva thankfully wau?

its just slowly getting better but its getting there none the less

Divide each term of the polynomial by 4

4x^4   - 25x^2  - 5x   - 13

___       _____   ___   ___      =   (1x^4) - ( 25/4)x^2  - (5/4)x - 13/4

 4             4         4        4

Write down the coefficients    =    1     - 25/4        -5/4      -13/4

Drop the first one and remove the minus signs from any of the rest  ....we get ... 25/4     5/4     13/4

We have two  possible bounds on the polynomial

a)  Take the largest value of the above and add 1  =   25/4 + 1 =  25/4 + 4/4   = 29/4  = 7.25

b)  Take the larger of :  the sum of the coefficients or 1....Add the coefficients = 43/4.....this is greater than 1....so the second possible bound is  43/4  = 10.75

Select the smaller  of  (a)  or (b)  =  7.25

This tells us that the upper and  lower bounds on zeroes are  7.25   and - 7.25

So.....the upper bound  is x = 7.25.....in other words......no real root will be  > 7.25